I'm motivated to ask this question because of the following question.
Consider the sequence $\{f_n\}$ defined on $[0, \pi]$ by $f_n(x) = \sin^n(x)$. Show that $\{f_n\}$ converges pointwise. Use the fact that if $\{f_n\}$ is a sequence of continuous functions that converges uniformly to $f$, then $f$ is continuous to show that $\{f_n\}$ is not uniformly convergent.
$f_n$ converges pointwise to $g(x)$ defined by $g(x) = 0$ if $x \neq \frac{\pi}{2}$, and otherwise $g(x) = 1$. Clearly, $g(x)$ is not continuous. Hence, $f$ cannot converge uniformly to $g$.
However, the question asked me to prove that $\{f_n\}$ is not uniformly convergent, but I have only proved that $f_n$ does not converge uniformly to $g$.
If indeed it was the case that $f_n \rightarrow f$ pointwise and $f_n \rightarrow f'$ uniformly, then $f = f'$, then I would be done since $f_n$ cannot converge uniformly to no other function than $g$.
But I'm not sure if this is true. Intuitively it seems true, but I might be missing something.
Thanks for the help in advance!